\(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)

\(\Leftrightarrow\left(\dfrac{x-3}{113}+1\right)+\left(\dfrac{x-5}{115}+1\right)=\left(\dfrac{x-7}{117}+1\right)+\left(\dfrac{x-9}{119}+1\right)\)\(\Leftrightarrow\dfrac{x+110}{113}+\dfrac{x+110}{115}=\dfrac{x+110}{117}+\dfrac{x+110}{119}\)

\(\Leftrightarrow\dfrac{x+110}{113}+\dfrac{x+110}{115}-\dfrac{x+110}{117}-\dfrac{x+110}{119}=0\)

\(\Leftrightarrow\left(x+110\right)\left(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\right)=0\)

Mà \(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\ne0\)

\(\Rightarrow x+110=0\)

\(\Rightarrow x=-110\)

\(\dfrac{x-3}{133}+\dfrac{x-5}{155}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)

\(\Leftrightarrow\left(\dfrac{x-3}{113}+1\right)+\left(\dfrac{x-5}{115}+1\right)=\left(\dfrac{x-7}{117}+1\right)+\left(\dfrac{x-9}{119}+1\right)\)

\(\Leftrightarrow\dfrac{x+130}{113}+\dfrac{x+130}{115}=\dfrac{x+130}{117}+\dfrac{x+130}{119}\)

\(\Leftrightarrow\dfrac{x+130}{113}+\dfrac{x+130}{115}-\dfrac{x+130}{117}-\dfrac{x+130}{119}=0\)

\(\Leftrightarrow\left(x+130\right)\left(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\right)=0\)

Mà \(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\ne0\)

\(\Leftrightarrow x+130=0\)

\(\Leftrightarrow x=-130\)

Vậy..

https://giaibaitapvenha.blogspot.com/2017/12/toan-lop-8-ai-so-giai-phuong-trinh.html

Nếu lời giải giúp ích được cho bạn, truy cập

https://giaibaitapvenha.blogspot.com/2017/12/en-voi-do-homework-for-you-e-trai.html

a,\(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)

=>\(\dfrac{x-3}{113}+1+\dfrac{x-5}{115}+1=\dfrac{x-7}{117}+1+\dfrac{x-9}{119}+1\)

=>\(\dfrac{x+110}{113}+\dfrac{x+110}{115}=\dfrac{x+110}{117}+\dfrac{x+110}{119}\)

=>(x+110)(\(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\))=0

vì 1/113+1/115-1/117-1/119 khác 0

=>x+110=0=>x=-110

\(\dfrac{x-291}{1700}+\dfrac{x-293}{1698}+\dfrac{x-295}{1696}+\dfrac{x-297}{1694}=4\)

=>\(\dfrac{x-291}{1700}-1+\dfrac{x-293}{1698}-1+\dfrac{x-295}{1696}-1+\dfrac{x-297}{1694}-1+4=4\)=>\(\dfrac{x-1991}{1700}+\dfrac{x-1991}{1698}+\dfrac{x-1991}{1696}+\dfrac{x-1991}{1694}=0\)

=>(x-1991)(1/1700+1/1698+1/1696+1/1694)=0

vì 1/1700+1/1698+1/1696+1/1694 khác 0

=>x-1991=0=>x=1991

Bài 1:

a) A = 2¹⁰+2¹¹+2¹² 

=2¹⁰*(1+2¹+2²)

=2¹⁰*(1+2+4)

=7*2¹⁰ chia hết 7

Đpcm

b)7*32=244

=32+64+128

=2⁵+2⁶+2⁷

Bài 2:

a)ko hiểu đề

b)nhân N với * x như dạng lp 6 âý

a, Đặt \(x=\frac{1}{117}\), \(y=\frac{1}{119}\) ta có:

\(A=\left(3+x\right)y-4x\left(5+1-y\right)-5xy+24x\)

\(=3y+xy-24x+4xy-5xy+24x\)

\(=3y\)

\(=\frac{3}{119}\)

b, Thay 8 bằng x + 1 ta có:\(B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...-x^3-x^2+x^2+x-5\)

\(=7-5\)

= 2

a) Đặt a = \(\frac{1}{117}\)và b = \(\frac{1}{119}\)

Theo đề ta có:

A = (3 + a) b - 4a ( 5+1-b)-5ab+24a

= 3b + ab - 20a -4a + 4ab - 5ab + 24a

= 3b

= 1.\(\frac{1}{119}\) = \(\frac{3}{119}\)

Vậy A = \(\frac{3}{119}\)

a) 75 x 48 - 9 x 90 + 6999

= 3600 - 810 + 6999

= 9789

~HT~

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